Math, asked by jeevanshud7, 7 months ago

if y=λx-3 ,y=μx+1 ,y=x+4 are 3 normals from point p to parabola whose axis is along x axis then 2λ-3μ is equal to

Answers

Answered by mohdaamir074
0

Answer:

kya h ye smjhe me hi ni a rha h

Answered by dualadmire
0

The value of 2λ - 3μ  is  - 5/2.

Given: The 3 normals of a parabola from a point are given as

y = λx - 3, y = μx + 1, y = x + 4.                                                            

To Find: The value of  2λ - 3μ.

Solution:

  • It is known that from a point, 3 normals can be drawn to a parabola. Such a point is known as a co-normal point.
  • The sum of the ordinates of the co-normal points is zero.
  • The ordinates (y) are given as, y = am

Where y = ordinate, a = constant, m = slope.

Coming to the numerical, the normals are given to us to are,

           y = λx - 3,                                                              ...(1)

           y = μx + 1,                                                              ...(2)

           y = x + 4.                                                                ...(3)

These are not in the form of y = am, so to convert we can,

      ⇒ y + 3 =  λx

      ⇒  Y = λx                                                                 ...(4)

      ⇒ y - 1 = μx

      ⇒  Y = μx                                                                 ...(5)

      ⇒ y - 4 = x

      ⇒   Y = x                                                                   ...(6)

According to bullet point 2, we can say that,

            ∑ Y = 0

          ⇒ y + 3 + y - 1 + y - 4 = 0

          ⇒ 3y = 2

          ⇒  y = 2/3                                                  

Putting y = 2/3 in y = x + 4, we get;

              y = x + 4

         ⇒  x = 2/3 - 4 = ( 2 - 12 ) / 3

         ⇒  x = - 10/3

Putting the respective values of x and y in (1), we get;

              y = λx - 3

          ⇒ 2/3 = λ × ( - 10/3 ) - 3

          ⇒  λ × ( - 10/3 ) = ( 2 + 9 ) / 3

          ⇒  λ  =  - 11 / 10                                                                      ...(7)

Putting the respective values of x and y in (2), we get;

             y = μx + 1

         ⇒ 2/3 = μ × ( - 10/3 ) + 1

         ⇒ μ × ( - 10/3 ) = ( 2 - 3 ) / 3

         ⇒  μ  =  1 / 10                                                                           ...(8)

Now, we need to find the value of the expression,

           2λ - 3μ

Putting the respective values of λ and μ from (7) and (8) in the expression, we get;

      ⇒ ( 2 × ( - 11/10 )) - ( 3 × 1/10 )

       ⇒  - 22/10 - 3/10

       ⇒ - 25/10

       ⇒ - 5 / 2

Hence, the value of 2λ - 3μ  is  - 5/2.

#SPJ2

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